Traffic Lights X-O-X

Traffic lights x-o-x is a 2-player game. You can play the game through the link below:

Enjoy the game!

Real Mathematics – Game #13

Crossing the bridge

Years and years ago, there was a village known as Togan. It was located in Mesopotamia’s fertile lands, next to rivers and beautiful waterfalls. There were only a few hundred people who lived in Togan. All of the Togan were farmers, except one: Berkut.


Berkut was the oldest man in the village. He had a long white beard. His story had become a kind of a legend within time. According to the people of Togan, whoever entered his property would never be seen again. This is why Berkut’s house was the only house that stayed on the other side of the river of Togan.

Berkut’s home.

For kids, Berkut was a mystery. Whenever Berkut was out on his garden, kids would gather and watch him from the other side of the river.

People of Togan were hardworking farmers. They would be working on their farms starting from their childhood. Like the rest of the Togan, Ali starting helping his family at an early age. Ali would work from sunrise to the sunset.

For Ali and his friends, Berkut’s situation was one of the hot topics. One day, these four friends decided that they would skip working and go across the river to investigate Berkut’s house. This legend had to be questioned!

The next day, while he was out on his garden, Berkut realized that four kids were about to cross the river, using the old bridge. He watched them crossing the bridge in pairs, as the old bridge was not strong enough to carry more than 2 of them at the same time.

He, then, went inside his house as Ali and his friends were approaching the house. When kids were inside the garden, they saw that a jar of cookies and a steamy teapot was waiting for them. While they were checking the table, Berkut went outside and greeted them. Kids, dumbfounded, started screaming as they all dispersed out of the garden in different ways.

They found each other only after it was dark. Now, kids, who had only one lamp with themselves, had to cross the bridge as fast as they can. But they couldn’t risk crossing the old bridge at once. Each time, only two of them could cross the bridge.

Crossing Times:
Jane: 1 minute
Ali: 2 minutes
Tom: 6 minutes
Jenny: 10 minutes

Since they had to cross the bridge in pairs while sharing a lamp, their speed would be at the rate of the slowest of the pair.

Now, you must solve this problem:

What is the fastest route to the other side of the river?

M. Serkan Kalaycıoğlu

Real MATHEMATICS – Game #12

Triangle Invasion

Invasion is a multiplayer game which needs only a pencil and a pen. To start the game one should draw a big triangle on a paper whilst assigning the corners with 1-2-3:


Then the big triangle should be triangulated (without any rules):


Once the construction is over, we can start labeling the triangle corners with the direction of the following rules:

  1. One can label 1-2 side with either of 1 or 2.
  2. One can label 1-3 side with either of 1 or 3.
  3. One can label 2-3 side with either of 2 or 3.
  4. One can label the inside of the original triangle with any of 1, 2 or 3.

Progress of the Game

  • Assign the corners according to the rules.
  • In order to invade a triangle, player must assign the final unattached corner of that specific triangle.
  • Goal is to avoid invading a triangle that has corners 1-2-3. Winner is the player who invaded least number of such triangles.

Sperner’s Triangle

Idea of the invasion game comes from the Sperner’s triangle which is discovered by Emanuel Sperner in the 20th century. Triangulated shape in the invasion game is an example of Sperner’s triangle.

After labeling the corners a Sperner triangle will always give a small triangle that has corners 1-2-3:


Actually, Sperner’s triangle always has odd-numbered 1-2-3 triangles.

This is why invasion game never ends with draw.

Dead End

Sperner’s triangle can be used to construct many games. For example, let’s say in a Sperner’s triangle all the sides of the little triangles that are connected between 1 and 2 are doors. And all the other sides are walls:

If one tries to walk through these doors, that person will end up in two situations:

  1. The person will end up inside a 1-2-3 triangle:
  2. The person will find him/herself outside of the triangle:

    M. Serkan Kalaycıoğlu

Real MATHEMATICS – Game #11


Sprouts is a multiplayer game which was created by M. S. Paterson and brilliant J. H. Conway back in 1967. All you need for playing sprouts are just a piece of paper and a pen/pencil.

  • Game starts with 3 dots on a paper:
  • Players take turns and draw lines from one dot to another (a line can be drawn to the same dot as well). Lines don’t have to be straight and a new dot must be placed on each line:
  • Lines can’t cross one another:
  • A dot is called “dead” if it has 3 lines coming out of it. In other words any dot can be connected to at most 3 lines:

    On right: A, B and E have 3 dots. This means A, B and E are all “dead”.

  • Player who draws the last possible line is the winner.

Brussels’ Sprouts

Brussels’ sprouts is a different kind of sprouts game. It is a multiplayer game just like regular sprouts and all it needs are a paper and a pen/pencil as well. But this time game starts with dots that have thorns. Assume that we will start a Brussels’ with two dots with 3 and 4 thorns on them:


Players take turn to draw lines between thorns. When a player draws a line, he/she should mark a new dot that has two thorns on it:

Just like regular sprouts, lines can’t cross in Brussels’ sprouts. And the player who draws the last line wins the game:

Euler and Sprouts

You might wonder how on Earth I get to mention Euler in a game that was created about 200 years after he passed away. I recommend you to check Euler characteristics article.

Euler says:

Let’s imagine that V dots and E lines (which don’t cross one another) are sitting on a plane. If the number of faces on this shape is F (don’t ever forget to count the whole plane as one face), then the equation

V – E + F = 2

will always be satisfied.

Take a finished Brussels’ sprouts game on hand:


Find the numbers of the dots, lines and faces:

Apply Euler’s formula:


Euler will always be right!

Four Colors

Now take any Brussels’ sprouts sheet and color the faces on it. (Neighboring faces have different colors.)

You will see that four different colors will be enough to color any Brussels’ sheet:

One wonders…

  1. At most how many turns can there be in a regular sprouts game that starts with 3 dots?
  2. Is there a winning strategy for sprouts?
  3. Start a Brussels’ sprouts with 3 dots. If each of them has 3 thorns, at most how many turns can there be?

M. Serkan Kalaycıoğlu

Real MATHEMATICS – Game #10

Free the Oreos

Free the Oreos is a multiplayer game for everyone aged over 6. Here is what you need for the game:

  • 9 identical squares drawn inside a bigger square:
  • 24 matches in order to construct those 9 little squares:
  • Placing an Oreo inside each little square:


  1. Players will remove matches in turns. A player will remove exactly one match whenever it is his/her turn.
  2. In order to free an Oreo inside a square, all the matches of that square must be removed:

    To free the Oreo sitting on top-left, all four sides of the square must be removed.
  3. The player who removed the last side of a square is known as the one who freed the Oreo.

Goal is to free more Oreos than your opponent.

Defense vs. Attack

Attacking Strategy: You should remove the last match surrounding the Oreo in case you want to free it.

Defending Strategy: In case there are two matches around a specific Oreo, you should avoid removing either of those matches. Otherwise your opponent can and will free that Oreo in his/her turn.

Two experienced players will use the defending strategy as long as possible. But in the end the game will be in a position where defending won’t be possible:


In such situation, game can continue as follows:

At the end of this sample game first player freed 6 Oreos as the second freed only 3.

One wonders…

  1. Can you find a winning strategy?
  2. At most how many Oreos can a player free in a single turn? Show your answer with a sample game.

M. Serkan Kalaycıoğlu

Real MATHEMATICS – Game #9

Worst Kind of Games

Football is the most popular game in the world right now. It has fairly simple rules which are easy to understand. In my opinion it is the greatest game ever. Although there is something which I despise in football: Draw. I really don’t feel right when a football match ends in draw. Nevertheless there is always only one champion which makes football great again.

Do they look like they will be happy to get a draw?!

A game ending with a draw is something I have issue with. I just can’t understand how a person can enjoy the possibility of a tie. I am sure some (or possibly most) of you disagree with me. I’ve mentioned about this in the previous posts: I am a gamer. I try to turn everything in my daily life into games, and I like to see a winner at all times. For example when I was a child I played tic-tac-toe literally thousands of times. Perhaps you played it too. What is strange is that I would make deliberate mistakes whenever I realized the possibility of the game ending in draw.

Solution is Hex

  • Make a diamond-shaped board out of nine regular hexagons as follows:
  • Let one of the opposite sides be red and the other blue:
  • One of the players becomes red team as other becomes blue team.
  • Throw a coin in order to choose who goes first.
  • Players can select any unoccupied hexagon.

Goal: Red player tries to build a red road between red sides as the blue one tries the same with blue color.

Hex has a significant difference from tic-tac-toe: There can never be a draw in Hex.

One of the players will always win the game as other one loses.

Example: Red goes first.

Let’s try to draw in Hex. This can only happen if there aren’t any connections between red-red and blue-blue.

Assume that red made his first two moves as follows:


In order to avoid a red bridge blue must play his/her second move like this:


Now red must prevent blue to have a connection. But there is a problem here as blue has two options for a win:


This is why no matter what red does, blue will win the game:

Our goal is to prevent a win. So let’s assume that blue is an amateur at Hex and he/she can’t see this possibility. In this case red will not only prevent the loss, he/she will win the game:

There is no escape from a win.


Let’s use randomness to show that there can’t be a draw in Hex. I numbered the hexagon boxes from 1 to 9 as follows:


Then I placed 5 red and 5 blue papers in a bag and started selecting papers one by one. I placed the first paper in the hexagon box number 1 and so on…

In the end I ended up with this:


As you can see the blue team won.

I tried the randomness one more time and ended up with red team winning:

Even when the play is random, there will be no draw.

Some History
Hex was first invented by Danish architect and mathematician Piet Hein in 1942. Although it is known that great mathematician John Nash found the same game independently from Hein in 1948. The name Hex was given by the Parker Brothers, the company that popularized Hex as a board game.

One wonders…

Original Hex board is in 11×11 sizes. You can click here for an original Hex board and play the game. Try to think of winning strategies for Hex.

M. Serkan Kalaycıoğlu

Real MATHEMATICS – Game #8

Judge Seesaw 2.0.1

In the previous post I asked how you can find the 12th box when we don’t know whether it is heavier or lighter than the rest of the boxes.

To start the solution, we will divide the boxes into three groups of 4. But this time I will assign names to these groups and boxes:

Eleven of them are identical. Which one is the oddly-weighted one?

In the first try we’ll place the groups A and B to the end of the seesaw. There will be two possible outcomes:

Outcome 1: Seesaw is in balance.


This outcome shows that the boxes of A and B are identical.

Then, we can conclude that oddly-weighted box is among the group C.

In the second try let’s place any three boxes of the group C with any three boxes of the group A. (Why three from A? Since all the boxes of A and B are identical, it doesn’t matter which three boxes I choose among them. I selected three from the group A. I could have selected two boxes from A and one from B as well. It wouldn’t change anything.)

After the second try, we will again left with two possible outcomes:

a. Seesaw is in balance.

C4 is the oddly-weighted one.

It means three boxes of C and A are identical. Then we can conclude that the forth box of C is the oddly-weighted box.

b. Seesaw is tilted.

This concludes that one of the three boxes I choose from C is the oddly-weighted box. In this case the seesaw is either tilted towards C or A which tells us our box is either heavier or lighter than the rest of the boxes.

In other words, after the second try we ended up with the following question: “Two of three boxes have the same weight. If you know whether the odd box is heavier or lighter than the other two, how can you find the odd box?”

This can be solved with exactly one try on the seesaw. Select any two of the three boxes, place them on the seesaw.

If seesaw is in balance, then the third box is the odd one:

C2 is the oddly-weighted one.

In case seesaw is tilted, as we know whether the odd box is heavier or lighter than the others, we can find the box we’ve been looking for:

Outcome 2: Seesaw is tilted.


In this case the box we are looking for is either inside the group A or inside the group B.

Another deduction we can make is that one of these groups is heavier than the other. For the sake of a clearer solution I am going to assume that A is heavier than B. (I could have chosen the other way around; it wouldn’t change anything.)

Before starting the second try I will be swapping a box between A and B. Also I will be replacing the remaining three boxes of B with three boxes of C.

There are three possible outcomes:

a. Seesaw is balanced.

Our box is either one of B2, B3 or B4

This means that all the boxes sitting on the seesaw are identical. Hence the box we are looking is among the three boxes of group B which we replaced. We can also deduce that oddly-weighted box is lighter than the other boxes since the seesaw is balanced.

At this point all we need to do is to find the lighter one inside three boxes. As we showed in the previous situation, it can be done with comparing any two of those boxes.

b. Seesaw is tilted in the same direction as the first try.

Our box is either one of A1, A2 or A3.

This means that the oddly-weighted box is among the three boxes of A. We can also conclude that this box is heavier than the rest.

Again we have three boxes with one of them heavier than the other two. We can find that box with simply comparing any two of those boxes with the seesaw.

c. Seesaw is tilted in the opposite direction of the first try.

Our box is either A4 or B1.

This means that the oddly-weighted box is either one of the boxes which we swapped between A and B. But in this situation we have no clue if the box we are looking for is heavier or lighter.

In order to find our box, we should compare one of those two boxes with any one of the remaining 10 boxes.

If there is a balance on the seesaw, it means our box is the one we didn’t place on the seesaw:

Our box is A4.

If the seesaw is tilted, then it means our box (in the following cases it is A4) is the one we’ve chosen and placed on the seesaw.

M. Serkan Kalaycıoğlu

Real MATHEMATICS – Game #7

Judge Seesaw

There are 12 gift boxes on the table. They all have identical sizes and gift wraps.

rectangular white and red gift box

11 of them weight the same as the 12th box weights different from others.

Goal: Finding the oddly-weighted box.

Prize: 50 million old Turkish liras worth Oreo. (Because Oreo is the most beautiful thing in life.)

Tools: You can use a seesaw to compare the weights of the boxes.


Challenge: You will have to guess the oddly-weighted box in at most 3 tries.

Judge Seesaw 1.0.1

It is the most basic version of the game where it is known if the 12th box is heavier or lighter than other 11 boxes.

At this point give yourself a few minutes and think on a solution when it is known that the 12th box is heavier than the rest.


Time is up. Here is the solution:

I assumed that the 12th box is heavier than the rest of the boxes. Then we should divide 12 boxes into three groups of 4.

In the first try we will take any two of those groups of 4 and place them on the ends of the seesaw. There will be exactly two possible outcomes:

Situation A: Seesaw is balanced.


Two groups have equal weights which would mean the oddly-weighted box must be among the third group of 4.


In the second try we will take the third group of 4, divide those two-by-two and place them on the ends of the seesaw.


We will be seeing that the seesaw is tilted to one side. That happens because heavy box sits on the seesaw.


In the third try we’ll be taking the heavier couple from previous measure and place them one-by-one on the ends of the seesaw.


The seesaw will be tilted in favor of the heavier box. Hence we found the oddly-weighted box in exactly three tries.


Situation B: Seesaw is tilted.



This happens because oddly-weighted box is among one of the groups of 4 boxes. Take that group and apply the same methods we did in situation A.

One wonders…

Judge Seesaw 2.0.1

In this version of the game we have no idea whether the 12th box is heavier or lighter than the rest of the boxes.

Even though it seems like a slight change, this game has become much harder compared to the version 1.0.1.

This is why I will give you a little time. I’ll include the answer in the next post.

M. Serkan Kalaycıoğlu